Carcass wrote:
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In the figure, RSTU is a square with a side of 4. Points V and W are the midpoints of sides \(\overline{UT}\) and \(\overline {ST}\), respectively. What is the area of \(\triangle RWV\), in square units?
A 5
B 6
C 7
D 8
E 9
In \(\triangle RUV\)
\(RV^2 = 4^2 + 2^2\)
\(RV = 2\sqrt{5}\)
In \(\triangle RSW\)
\(RW^2 = 4^2 + 2^2\)
\(RW = 2\sqrt{5}\)
In \(\triangle WTV\)
\(WV^2 = 2^2 + 2^2\)
\(WV = 2\sqrt{2}\)
Now, we can see that \(\triangle RWV\) is an isosceles triangle. We can apply either Heron's formula or drop a perpendicular from point R to base VW (at X)
\((2\sqrt{5})^2 = RX^2 + (\sqrt{2})^2\)
\(RX = \sqrt{(20 - 2)} = \sqrt{18} = 3\sqrt{2}\)
Area of \(\triangle RWV\) = \(\frac{1}{2}(2\sqrt{2})(3\sqrt{2}) = \frac{1}{2}(12) = 6\)
Hence, option B
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GRE In the figure, RSTU is a square with a side of 4.jpg [ 9.52 KiB | Viewed 2691 times ]